Cyclpropane is converted to its isomer propylene, when heated. The rate law is first order in cyclopropane and the rate constant is 6.0X10^-4/s at 500 degrees C. If the intial concentration of cyclopropane is 0.0226 mol/L, what is the concentration after 899s?
Isn't the first order equation
ln(Ao/A) = kt
Then ln(0.0226/A) = kt.
Substitute the k value and time of 899 seconds, and solve for A. Check my thinking.
To determine the concentration of cyclopropane after 899 seconds, we need to use the first-order rate law equation:
ln(c2/c1) = -kt
Where:
ln(c2/c1) is the natural logarithm of the ratio of concentrations
c1 is the initial concentration of cyclopropane
c2 is the concentration of cyclopropane after a certain time
k is the rate constant
t is the time interval
Given data:
Initial concentration (c1) = 0.0226 mol/L
Rate constant (k) = 6.0 × 10^-4/s
Time interval (t) = 899 seconds
Now, let's plug the values into the equation to find the concentration of cyclopropane after 899 seconds.
ln(c2/0.0226) = - (6.0 × 10^-4/s) × (899 s)
Rearranging the equation to solve for c2:
ln(c2/0.0226) = - (6.0 × 10^-4/s) × (899 s)
Using the property of logarithms, we can rewrite the equation as:
c2/0.0226 = e^(-(6.0 × 10^-4/s) × (899 s))
Now we can solve for c2 by multiplying both sides by 0.0226:
c2 = 0.0226 × e^(-(6.0 × 10^-4/s) × (899 s))
Using a calculator, we can evaluate the exponential term:
c2 ≈ 0.0226 × e^(-0.5364)
c2 ≈ 0.0226 × 0.5849
c2 ≈ 0.0132 mol/L
Therefore, the concentration of cyclopropane after 899 seconds is approximately 0.0132 mol/L.